I am solving a differential equation d y d t = y + 1 t + 1 .I got the solution y = c ( t + 1 ) − 1, c a constant.But the handout by my professor says"The solution is y = c ( t + 1 ) − 1, c a constant. where t ≠ − 1. But it does not mean that y is not defined at t ≠ − 1. It means that y can take any value at t = − 1 as long as it satisfies y = c ( t + 1 ) − 1."It does not make much sense to me because if y satisfies the equation y = c ( t + 1 ) − 1 at t = − 1, then the solution will be just y = c ( t + 1 ) − 1, c a constant.And I believe my solution indeed makes sense because y = c ( t + 1 ) − 1, c a constant is defined on R and differentiable, y ′ = c for any t ∈ R . On the other hand, plugging this in to y + 1 t + 1 gives me c for any t ∈ R .Would you explain what the handout is saying?

Question

I am solving a differential equation                     d        y                    d        t              =                  y        +        1                    t        +        1            .I got the solution   y  =  c  (  t  +  1  )  −  1,   c a constant.But the handout by my professor says&quot;The solution is   y  =  c  (  t  +  1  )  −  1,   c a constant. where   t  ≠  −  1. But it does not mean that   y is not defined at   t  ≠  −  1. It means that y can take any value at   t  =  −  1 as long as it satisfies   y  =  c  (  t  +  1  )  −  1.&quot;It does not make much sense to me because if   y satisfies the equation   y  =  c  (  t  +  1  )  −  1 at   t  =  −  1, then the solution will be just   y  =  c  (  t  +  1  )  −  1,   c a constant.And I believe my solution indeed makes sense because   y  =  c  (  t  +  1  )  −  1,   c a constant is defined on       R   and differentiable,       y    ′    =  c for any   t  ∈      R  . On the other hand, plugging this in to                     y        +        1                    t        +        1             gives me   c for any   t  ∈      R  .Would you explain what the handout is saying?

Jaida Sanders · Accepted Answer

Since the derivative             d      y              d      t       is not defined at   t  =  −  1 then all that can be found is solutions of the differential equation on the intervals   (  −  ∞  ,  −  1  ) and   (  −  1  ,  +  ∞  ) separately.

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